Question

A sample of hydrogen gas at 341 K experiences a change in volume from 5.49 L to 3.12 L. If its new pressure is 3.48 atm at 215 K, what was its original pressure in atmospheres?

Answer 1

Q172251

A sample of hydrogen gas at 341 K experiences a change in volume from 5.49 L to 3.12 L. If its new pressure is 3.48 atm at 215 K, what was its original pressure in atmospheres?'

Solution:

Initial temperature of hydrogen gas is 341K.

Volume of sample of hydrogen gas changed from 5.49 L to 3.12L, so

V _initial = 5.49 L and V _final = 3.12 L.

New pressure is 3.48 atm at 215K, which means

P _final = 3.48 atm and T _final = 215 K.

We have to find the original pressure ( initial ) in 'atm',

P _initial = unknown ; P _final = 3.48 atm

V _initial = 5.49 L ; V _final = 3.12 L

T _initial = 341 K ; T _final = 215 K

We can use the combined gas equation here.

$$\frac{P_1\space V_1}{T_1} = \frac{P_2\space V_2}{T_2}$$

$$\frac{P_{initial}\space V_{initial}}{T_{initial}} = \frac{P_{final }\space V_{final }}{T_{final }}$$

Substitute all the given information in the formula, we have

$$\frac{P_{initial}\space *\space 5.49\space L}{341\space K} = \frac{3.48\space atm\space *\space 3.12\space L }{215\space K }$$

$$P_{initial}\space * 0.01610 L/K = 0.050500L.atm/K$$

divide both the side by 0.01610 L/K, we have

$$\frac{P_{initial}\space *\space 0.01610\space L/K}{0.01610\space L/K} = \frac{0.050500\space L.atm/K}{0.01610\space L/K }$$

$$P_{initial} = \frac{0.050500\space L.atm/K}{0.01610\space L/K } = 3.137 atm.$$

In the question all the quantities are given in 3 significant figure, so our final answer must also

be in 3 significant figure.

Hence the Original (initial) pressure of sample of hydrogen gas will be 3.14 atm.